Evaluation of Some Non-Elementary Integrals Involving Sine, Cosine, Exponential and Logarithmic Integrals: Part I

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URAL MATHEMATICAL JOURNAL, 2018, Vol. 4, No. 1, pp. 24–42 DOI: 10.15826/umj.2018.1.003

EVALUATION OF SOME NON-ELEMENTARY INTEGRALS INVOLVING SINE, COSINE, EXPONENTIAL AND LOGARITHMIC INTEGRALS: PART I

Victor Nijimbere

School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada [email protected]

β α Abstract: The non-elementary integrals Siβ,α = R [sin(λx )/(λx )]dx, β ≥ 1, α ≤ β +1 and Ciβ,α = R [cos (λxβ )/(λxα)]dx, β ≥ 1, α ≤ 2β + 1, where {β, α} ∈ R, are evaluated in terms of the hypergeometric functions 1F2 and 2F3, and their asymptotic expressions for |x|≫ 1 are also derived. The integrals of the form n β α n β α R [sin (λx )/(λx )]dx and R [cos (λx )/(λx )]dx, where n is a positive integer, are expressed in terms Siβ,α and Ciβ,α, and then evaluated. Siβ,α and Ciβ,α are also evaluated in terms of the hypergeometric function 2F2. And so, the hypergeometric functions, 1F2 and 2F3, are expressed in terms of 2F2. The exponential integral λxβ α x Eiβ,α = R (e /x )dx where β ≥ 1 and α ≤ β + 1 and the logarithmic integral Li = Rµ dt/ ln t, µ > 1, are also expressed in terms of 2F2, and their asymptotic expressions are investigated. For instance, it is found that for x ≫ 2, Li ∼ x/ln x + ln(ln x/ln 2) − 2 − ln 2 2F2(1, 1; 2, 2; ln 2), where the term ln (ln x/ln2) − 2 − ln 2 2F2(1, 1; 2, 2; ln 2) is added to the known expression in mathematical literature Li ∼ x/ln x. The method used in this paper consists of expanding the integrand as a Taylor and integrating the series term by term, and can be used to evaluate the other cases which are not considered here. This work is motivated by the applications of sine, cosine exponential and logarithmic integrals in Science and Engineering, and some applications are given. Key words: Non-elementary integrals, Sine integral, Cosine integral, Exponential integral, Logarithmic integral, Hyperbolic sine integral, Hyperbolic cosine integral, Hypergeometric functions, Asymptotic evaluation, Fundamental theorem of calculus.

1. Introduction

Definition 1. An elementary function is a function of one variable constructed using that variable and constants, and by performing a finite number of repeated algebraic operations involving exponentials and logarithms. An indefinite integral which can be expressed in terms of elementary functions is an elementary integral. And if, on the other hand, it cannot be evaluated in terms of elementary functions, then it is non-elementary [6, 10].

Liouville 1938’s Theorem gives conditions to determine whether a given integral is elementary or non-elementary [6, 10]. For instance, it was shown in [6, 10], using Liouville 1938’s The- orem, that the integral Si1,1 = (sin x/x)dx is non-elementary. With similar arguments as in [6, 10], One can show that Ci = (cos x/x)dx is also non-elementary. Using the Euler formulas 1,1 R e±ix = cos x i sin x, and noticing that if the integral of a function g(x) is elementary, then both ± R its real and imaginary parts are elementary [6], one can, for instance, prove that the integrals Si = [sin (λxβ)/(λxα)]dx, β 1, α 1, and Ci = [cos (λxβ)/(λxα)]dx, where β 1 and β,α ≥ ≥ β,α ≥ α 1, are non-elementary by using the fact that their real and imaginary parts are non-elementary. ≥ R R The integrals [sinn (λxβ)/(λxα)]dx and [cosn (λxβ)/(λxα)]dx, where n is a positive integer, are also non-elementary since they can be expressed in terms of Si and Ci . R R β,α β,α To my knowledge, no one has evaluated these integrals before. To this end, in this paper, formulas for these non-elementary integrals are expressed in terms of the hypergeometric functions 1F2 and F whose properties, for example, the asymptotic expansions for large argument ( λx 1), 2 3 | |≫ Some non-elementary integrals of sine, cosine and exponential integrals type 25 are known [9]. We do so by expanding the integrand in terms of its Taylor series and by integrating the series term by term as in [7]. And therefore, their corresponding deﬁnite integrals can be evaluated using the Fundamental Theorem of Calculus (FTC). For example, the sine integral

B sin (λxβ) Si = dx, β 1, α β + 1, β,α (λxα) ≥ ≤ AZ is evaluated for any A and B using the FTC. λxβ α On the other hand, the integrals Eiβ,α = (e /x )dx and dx/ ln x, are expressed in terms of the hypergeometric function F . This is quite important since one may re-investigate the 2 2 R R asymptotic behavior of the exponential (Ei) and logarithmic (Li) integrals [3] using the asymptotic expressions of the hypergeometric function 2F2 which are known [9]. Some other non-elementary integrals which can be written in terms of Eiβ,α or dx/ ln x are βx also evaluated. For instance, as a result of substitution, the integral eλe dx is written in terms β R of Ei = (eλx /x)dx and then evaluated in terms of F , and using integration by parts, the β,1 2 2 R integral ln(ln x)dx is written in terms of dx/ ln x and then evaluated in terms of F as well. R 2 2 Using the Euler identity e±ix = cos(x) i sin(x) or the hyperbolic identity e±x = cosh(x) R R± ± sinh(x), Siβ,α and Ciβ,α are evaluated in terms Eiβ,α. And hence, the hypergeometric functions 1F2 and 2F3 are expressed in terms of the hypergeometric 2F2. This type of integrals find applications in many fields in Science and Engineering. For instance, in wireless telecommunications, the random attenuation capacity of a channel, known as fading capacity, is calculated as [11] ∞ 1 1 C = E[log (1 + P H 2)] = log (1 + Pξ)e−ξdξ = e1/P E ( ) E , fading 2 | | 2 ln 2 1,1 ∞ − 1,1 P Z0 where the fading coefficient H is a complex Gaussian random variable, and E X 2 P is the | | ≤ maximum average transmitted power of a complex-valued channel input X. In number theory, the prime number theorem states that [3] x dx π(x) Li(x)= , µ> 1, ∼ ln x Zµ where π(x) denotes the number of primes small than or equal to x. Moreover, there are applications of sine and cosine integrals in electromagnetic theory, see for example Lebedev [5]. Therefore, it is quite important to adequately evaluate these integrals. For that reason, the main goal of this paper is to evaluate non-elementary integrals of sine, cosine, exponential and logarithmic integrals type in terms of elementary and special functions with well known properties so that the fundamental theorem of calculus can be used so that we can avoid to use numerical integration. β α Part I is indeed devoted to the cases Siβ,α = [sin (λx )/(λx )]dx, β 1, α β + 1, β α λxβ α≥ ≤ Ciβ,α = [cos (λx )/(λx )]dx, β 1, α 2β +1 and Eiβ,α = (e /x )dx where β 1, ≥ ≤ R β α ≥ α β + 1, where β, α R. The other cases Siβ,α = [sin (λx )/(λx )]dx, β 1, α>β + 1, ≤ R { } ∈ R β ≥ Ci = [cos (λxβ)/(λxα)]dx, β 1, α> 2β +1 and Ei = (eλx /xα)dx where β 1, β,α ≥ R β,α ≥ α > β + 1, where β, α R, which may involve series whose properties are not necessary known R { } ∈ R will be considered in Part 2 [8]. Before we proceed to the objectives of this paper (see sections 2, 3, 4 and 5), we first define the generalized hypogeometric function as it is an important mathematical that we are going to use throughout the paper. 26 Victor Nijimbere

Deﬁnition 2. The generalized hypergeometric function, denoted as pFq, is a special function given by the series [1, 9]

∞ (a ) (a ) (a ) xn F (a , a , , a ; b , b , , b ; x)= 1 n 2 n · · · p n , p q 1 2 · · · p 1 2 · · · q (b ) (b ) (b ) n! n=0 1 n 2 n q n X · · · where a , a , , a and ; b , b , , b are arbitrary constants, (ϑ) = Γ(ϑ + n)/Γ(ϑ) (Pochham- 1 2 · · · p 1 2 · · · q n mer’s notation [1]) for any complex ϑ, with (ϑ)0 = 1, and Γ is the standard gamma function [1, 9].

2. Evaluation of the sine integral and related integrals

1 3 3 λ2x2 Proposition 1. The function G(x)= x F ; , ; , where F is a hypergeometric 1 2 2 2 2 − 4 1 2 sin (λx) function [1] and λ is an arbitrarily constant, is the antiderivative of the function g(x)= . λx Thus, sin (λx) 1 3 3 λ2x2 dx = x F ; , ; + C. λx 1 2 2 2 2 − 4 Z

P r o o f. To prove Proposition 1, we expand g(x) as Taylor series and integrate the series term by term. We also use the gamma duplication formula [1]

1 1 1 Γ(2α) = (2π)− 2 22α− 2 Γ(α)Γ α + , α C, 2 ∈ the Pochhammer’s notation for the gamma function [1],

Γ(α + n) (α) = α(α + 1) (α + n 1) = , α C, n · · · − Γ(α) ∈ and the property of the gamma function Γ(α+1) = αΓ(α) (eg., Γ (n + 3/2) = (n + 1/2) Γ (n + 1/2) for any real n). We then obtain

∞ ∞ sin (λx) 1 (λx)2n+1 λ2n x2n+1 g(x)dx= dx= ( 1)n dx = λ ( 1)n + C λx λx − (2n + 1)! − (2n + 1)! 2n + 1 Z Z Z n=0 n=0 ∞ X ∞ X x λ2n x2n x Γ (n + 1/2) = ( 1)n + C = ( λ2x2)n + C 2 − (2n + 1)! n + 1/2 2 Γ(2n + 2)Γ (n + 3/2) − n=0 n=0 ∞X n X (1/2) λ2x2/4 1 3 3 λ2x2 = x n − + C = x F ; , ; + C = G(x)+ C. (3/2) (3/2) n! 1 2 2 2 2 − 4 n=0 n n X

In the following lemma, we assume that the function G(x) is unknown and therefore we establish its properties such as the inﬂection points and its behaviour as x . → ±∞ sin x Lemma 1. Let G(x) be the antiderivative for g(x)= (λ = 1), and G(0) = 0. x Some non-elementary integrals of sine, cosine and exponential integrals type 27

1. Then G(x) is linear around x = 0 and the point (0, G(0)) = (0, 0) is an inﬂection point of the curve Y = G(x), x R. ∈ 2. And lim G(x)= θ while lim G(x)= θ, where θ is a positive ﬁnite constant. x→−∞ − x→+∞

Proof.

1. The series ∞ sin x (λx)2n g(x)= = ( 1)n x − (2n + 1)! nX=0 gives G′(0) = g(0) = 1. Then, around x = 0, G(x) x since G′(0) = g(0) = 1 and G(0) = 0. ∼ Moreover, ∞ sin x ′ (2n + 2)(λx)2n G′′(x)= g′(x)= = λ2x ( 1)n , x − − (2n + 3)! n=0 X and so G′′(0) = g′(0) = 0. Hence, by the second derivative test, the point (0, G(0)) = (0, 0) is an inﬂection point of the curve Y = G(x).

2. It is straight forward, using Squeeze theorem, to obtain lim g(x) = lim g(x) = 0. And x→−∞ x→+∞ since both g(x) and G(x) are analytic on R, then G(x) has to be constant as x by → ±∞ Liouville Theorem (section 3.1.3 in [4]) since if a complex function is entire on C then both its imaginary and real parts are analytic on the real line R including at x . Also, → ±∞ there exists some numbers δ > 0 and ǫ such that if x > δ then sin x /x 1/x < ǫ, and | | || | − | lim ( sin x /x)/(1/x) = lim ( sin x /x)/(1/x) = 1. This makes the function g1(x) = x→−∞ | | x→+∞ | | ± 1/x an envelop of g(x) away from x = 0 if sin x < 0 and g (x) = 1/x an envelop of g(x) − 2 away from x =0 if sin x> 0. Moreover, on one hand, g′ G′′ g′ if x< δ, and g′ and g′ 2 ≤ ≤ 1 − 1 2 do not change signs. While on another hand, g′ G′′ g′ if x > δ, and also g′ and g′ do 1 ≤ ≤ 2 1 2 not change signs. Therefore there exists some number θ > 0 such G(x) oscillates about θ if x > δ and G(x) oscillates about θ if x< δ. And G(x) θ if x δ. − − | |≤ | |≤ Example 1. For instance, if λ = 1, then

sin x 1 3 3 x2 dx = x F ; , ; + C. (2.1) x 1 2 2 2 2 − 4 Z sin x 1 3 3 x2 By Proposition 1, the antiderivative of g(x) = is G(x) = x F ; , ; , and the x 1 2 2 2 2 − 4 graph of G(x) is shown in Figure 1. It is in agreement with Lemma 1. It is seen in Figure 1 that (0, G(0)) = (0, 0) is an inﬂection point and that G attains some constants as x as predicted → ±∞ by Lemma 1.

In the following lemma, we obtain the values of G(x), the antiderivative of the function g(x) = sin (λx)/(λx), as x using the asymptotic expansion of the hypergeometric func- → ±∞ tion 1F2.

Lemma 2. Consider G(x) in Proposition 1,and preferably assume that λ> 0. 28 Victor Nijimbere

2 θ = π 1.5 2

0.5

G(x) 0

−0.5

−1

−1.5 θ = π 2 −2 − − −25 −20 −15 −10 −5 0 5 10 15 20 25 x

Figure 1. G(x) is the antiderivative of sin(x)/x given in (2.1).

1. Then, 1 3 3 λ2x2 π G( ) = lim G(x) = lim x 1F2 ; , ; = , (2.2) −∞ x→−∞ x→∞ 2 2 2 − 4 −2λ and 1 3 3 λ2x2 π G(+ ) = lim G(x) = lim x 1F2 ; , ; = . (2.3) ∞ x→+∞ x→∞ 2 2 2 − 4 2λ 2. And by the FTC, ∞ sin (λx) π √π π dx = G(+ ) G( )= = . (2.4) λx ∞ − −∞ 2λ − − 2λ λ −∞Z

Proof.

1. To prove (2.2) and (2.3), we use the asymptotic formula for the hypergeometric function 1F2 which is valid for z 1 and π arg z π, where arg z is the argument of z in the | | ≫ − ≤ ≤ complex plane. It can be derived using formulas (16.11.1), (16.11.2) and (16.11.8) in [9] and is given by F (a ; b , b ; z)= 1 2 1 1 2 − R−1 −n −a1 (a1)n ( z) −R Γ(b1)Γ(b2)z − + O( z ) ( Γ(b1 a1 n)Γ(b2 a1 n) n! | | ) Xn=0 − − − − 1/2 −iπ/2 S−1 Γ(b )Γ(b ) e2z e (ze−iπ)(a1−b1−b2+1/2)/2 µ (2.5) + 1 2 + n (ze−iπ)−n + O( z −S ) Γ(a ) √π 2n+1 | | 1 (n=0 ) 1 2 2 X 2z / eiπ/ iπ (a1−b1−b2+1/2)/2 S−1 Γ(b1)Γ(b2) e (ze ) µn iπ −n −S + n+1 (ze ) + O( z ) , Γ(a1) √π ( 2 | | ) nX=0 where a1, b1 and b2 are constants and the coeﬃcient µn is given by formula (16.11.4) in [9]. 2 2 We then set z = λ x /4, a1 = 1/2, b1 = 3/2 and b2 = 3/2, and we obtain

2 2 R−1 −2n −2R 1 3 3 λ x π −1/2 (1/2) λx λx F ; , ; = λ2x2 n i + O 1 2 2 2 2 − 4 2 n! 2 2 (n=0 !) X

Some non-elementary integrals of sine, cosine and exponential integrals type 29

−iλx S−1 −2n −2S √π e µn λx λx 2 2 n i + O −λ x 2 ( 2 − 2 2 !) Xn=0 S−1 √π eiλx µ λx −2n λx − 2S n i + O . −λ2x2 2 2n 2 2 (n=0 !) X

Then, for x 1, | |≫ R−1 −2n −2R π −1/2 (1/2) λx λx π λ2x2 n i + O , 2 n! 2 2 ∼ 2λ x (n=0 !) | | X

while S−1 S−1 √π eiλx µ λx −2n µ λx −2n λx −2S n i n i + O −λ2x2 2 2n − 2 − 2n 2 2 (n=0 n=0 !) X X √π eiλx + e−iλx cos (λx) = √π . ∼ (λx)2 2 (λx)2 We then obtain,

1 3 3 λ2x2 π x √π cos (λx) x F ; , ; , x . 1 2 2 2 2 − 4 ∼ 2λ x − λ λx | |→∞ | | Hence,

1 3 3 λ2x2 π x √π cos (λx) π G( ) = lim x1F2 ; , ; = lim = −∞ x→−∞ 2 2 2 − 4 x→−∞ 2λ x − λ λx −2λ | | and 1 3 3 λ2x2 π x √π cos (λx) π G(+ ) = lim x1F2 ; , ; = lim = . ∞ x→+∞ 2 2 2 − 4 x→+∞ 2λ x − λ λx 2λ | | 2. By the FTC,

+∞ 0 y sin (λx) sin (λx) sin (λx) dx = lim dx + lim dx = G(+ ) G( ) λx y→−∞ λx y→+∞ λx ∞ − −∞ −∞Z Zy Z0 1 3 3 λ2y2 1 3 3 λ2y2 π = lim y 1F2 ; , ; lim y 1F2 ; , ; = . y→+∞ 2 2 2 − 4 − y→−∞ 2 2 2 − 4 λ We now verify whether this is correct or not using Fubini’s Theorem [2]. We ﬁrst observe that

+∞ +∞ sin (λx) sin (λx) dx = 2 dx λx λx −∞Z Z0 since the integrand is an even function. We have in terms of double integrals and using Fubini’s Theorem that

+∞ +∞ +∞ +∞ +∞ sin (λx) 1 dx = e−sxsin (λx)dsdx = e−sxsin (λx)dxds. (2.6) λx λ Z0 Z0 Z0 Z0 Z0 30 Victor Nijimbere

Now using the fact that the inside integral in (2.6) is the Laplace transform of sin (λx) [1] yields

+∞ +∞ +∞ λ π e−sxsin (λx)dxds = ds = arctan (+ ) arctan 0 = . s2 + λ2 ∞ − 2 Z0 Z0 Z0

Hence, +∞ +∞ +∞ +∞ sin (λx) sin (λx) 2 π π dx = 2 dx = e−sxsin (λx)dsdx = 2 = λx λx λ 2λ λ −∞Z Z0 Z0 Z0 as before.

Setting λ = 1 as in Example 1, Lemma 2 gives lim G(x) = θ = π/2 while lim G(x) = x→−∞ − − x→+∞ θ = π/2. And these are the exact values of G(x) as x in Figure 1. → ±∞ Theorem 1. If β 1 and α < β + 1, then the function ≥ xβ−α+1 α 1 1 α 1 3 3 λ2x2β G(x)= F + + ; + + , ; , β α + 1 1 2 −2β 2β 2 −2β 2β 2 2 − 4 − where 1F2 is a hypergeometric function [1] and λ is an arbitrarily constant, is the antiderivative of sin (λxβ) the function g(x)= . Thus, λxα

sin (λxβ) xβ−α+1 α 1 1 α 1 3 3 λ2x2β Si = dx = F + + ; + + , ; + C. β,α λxα β α + 1 1 2 −2β 2β 2 −2β 2β 2 2 − 4 Z − (2.7)

And for x 1, | |≫ xβ−α+1 α 1 1 α 1 3 3 λ2x2β F + + ; + + , ; β α + 1 1 2 −2β 2β 2 −2β 2β 2 2 − 4 − (2.8) (2/λ)1+1/β−α/β Γ ( α/(2β) + 1/(2β) + 3/2) √π xβ−α+1 β α + 1 √π cos λxβ − − . ∼ β α + 1 Γ(1+ α/(2β) 1/(2β)) 2 x β−α+1 − β λ2 xβ+α−1 − − | |

Proof.

∞ sin (λxβ) 1 (λxβ)2n+1 Si = g(x)dx = dx = ( 1)n dx β,α λxα λxα − (2n + 1)! Z Z Z n=0 ∞ ∞ X λ2n λ2n = ( 1)n x2βn+β−αdx = λ ( 1)n x2βn+β−αdx − (2n + 1)! − (2n + 1)! n=0 Z n=0 Z X ∞ X λ2n x2βn+β−α+1 = ( 1)n + C − (2n + 1)! 2βn + β α + 1 n=0 − ∞X xβ−α+1 λ2n x2βn = ( 1)n + C 2β − (2n + 1)! n α/(2β) + 1/(2β) + 1/2 n=0 X − Some non-elementary integrals of sine, cosine and exponential integrals type 31

∞ xβ−α+1 Γ (n α/(2β) + 1/(2β) + 1/2) = − ( λ2x2β)n + C 2β Γ(2n + 2)Γ (n α/(2β) + 1/(2β) + 3/2) − n=0 − X∞ n xβ−α+1 ( α/(2β) + 1/(2β) + 1/2) λ2x2β/4 = − n − + C β α + 1 (3/2) ( α/(2β) + 1/(2β) + 3/2) n! n=0 n n − X − xβ−α+1 α 1 1 α 1 3 3 λ2x2β = F + + ; + + , ; + C = G(x)+ C. β α + 1 1 2 −2β 2β 2 −2β 2β 2 2 − 4 −

To prove (2.8), we use the asymptotic formula for the hypergeometric function 1F2 in equation (2.5), and proceed as in Lemma 2.

Beside, we can show as above that if β 1 and α < β + 1, then ≥ sinh (λxβ) xβ−α+1 α 1 1 α 1 3 3 λ2x2β dx = F + + ; + + , ; + C. (2.9) λxα β α + 1 1 2 −2β 2β 2 −2β 2β 2 2 4 Z −

Corollary 1. Let β = α. If α 1, then ≥

0 sin (λxα) 2 1/α Γ (1/(2α) + 1) √π dx = G(0) G( )= , (2.10) λxα − −∞ λ Γ (3/2 1/(2α)) 2 −∞Z −

+∞ sin (λxα) 2 1/α Γ (1/(2α) + 1) √π dx = G(+ ) G(0) = (2.11) λxα ∞ − λ Γ (3/2 1/(2α)) 2 Z0 − and +∞ sin (λxα) 2 1/α Γ (1/(2α) + 1) dx = G(+ ) G( )= √π. (2.12) λxα ∞ − −∞ λ Γ (3/2 1/(2α)) −∞Z −

Proof. If β = α, Theorem 1 gives

1 1 3 λ2x2α G( ) = lim x1F2 ; + 1, ; −∞ x→−∞ 2α 2α 2 − 4 2 1/α √π Γ (1/(2α) + 1) x √π cos (λxα) 2 1/α Γ (1/(2α) + 1) √π = lim = x→−∞ λ 2 Γ (3/2 1/(2α)) x − αλ2 x2α−1 − λ Γ (3/2 1/(2α)) 2 − | | ! − and

1 1 3 λ2x2α G(+ ) = lim x1F2 ; + 1, ; ∞ x→+∞ 2α 2α 2 − 4 2 1/α √π Γ (1/(2α) + 1) x √π cos (λxα) 2 1/α Γ (1/(2α) + 1) √π = lim = . x→+∞ λ 2 Γ (3/2 1/(2α)) x − αλ2 x2α−1 λ Γ (3/2 1/(2α)) 2 − | | ! − 32 Victor Nijimbere

Hence, by the FTC, 0 sin (λxα) 2 1/α Γ (1/(2α) + 1) √π α dx = G(0) G( ) = 0 λx − −∞ − − λ Γ (3/2 1/(2α)) 2 ! −∞Z − 2 1/α Γ (1/(2α) + 1) √π = , (2.13) λ Γ (3/2 1/(2α)) 2 − +∞ sin (λxα) 2 1/α Γ (1/(2α) + 1) √π dx = G(+ ) G(0) = 0 λxα ∞ − λ Γ (3/2 1/(2α)) 2 − Z0 − 2 1/α Γ (1/(2α) + 1) √π = . (2.14) λ Γ (3/2 1/(2α)) 2 − And combining (2.13) and (2.14) gives (2.12). Theorem 2. If β 1 and α < β + 1, then the FTC gives ≥ B sin (λxβ) dx = G(B) G(A), (2.15) λxα − AZ for any A and any B, and where G is given in Theorem 1. P r o o f. Equation (2.15) holds by Theorem 1, Corollary 1 and Lemma 2. Since the FTC works for A = and B = 0 in (2.10), A = 0 and B =+ in (2.11) and A = and B =+ −∞ ∞ −∞ ∞ in (2.12) by Corollary 1 for any β = α 1 and by Lemma 2 for β = α = 1, then it has to work ≥ for other values of A, B R and for β 1 and α < β + 1 since the case with β = α 1 is derived ∈ ≥ ≥ from the case with β 1 and α < β + 1. ≥ Theorem 3. Let β 1, then the function ≥ 2 β λx /2 5 λ2x2β G(x)=ln x F 1, 1; 2, 2, ; , | |− 6β 2 3 2 − 4 where 2F3 is a hypergeometric function [1] and λ is an arbitrarily constant, is the antiderivative of sin (λxβ) the function g(x)= . Thus, λxβ+1 2 sin (λxβ) λxβ/2 5 λ2x2β Si = dx = ln x F 1, 1; 2, 2, ; + C. β,β+1 λxβ+1 | |− 6β 2 3 2 − 4 Z

Proof. ∞ sin (λxβ) 1 (λxβ)2n+1 Si = g(x)dx = dx = ( 1)n dx β,β+1 λxβ+1 λxβ+1 − (2n + 1)! Z Z Z n=0 ∞ ∞ X λ2n dx λ2n = ( 1)n x2βn−1dx = + ( 1)n x2βn−1dx − (2n + 1)! x − (2n + 1)! n=0 Z Z n=1 Z X ∞ X λ2n+2 x2βn+2β = ln x + ( 1)n+1 + C | | − (2n + 3)! 2βn + 2β n=0 X ∞ λ2x2β λ2n x2βn = ln x ( 1)n + C | |− 2β − (2n + 3)! n + 1 n=0 X Some non-elementary integrals of sine, cosine and exponential integrals type 33

∞ n λ2x2β (Γ (n + 1))2 λ2x2β = ln x − + C | |− 2β Γ(2n + 4)Γ (n + 2) n! n=0 X2 ∞ n λxβ/2 (1) (1) λ2x2β/4 = ln x n n − + C | |− 6β (2) (2) (5/2) n! n=0 n n n 2 X λxβ/2 5 λ2x2β = ln x F 1, 1; 2, 2, ; + C = G(x)+ C. | |− 6β 2 3 2 − 4

3. Evaluation of the cosine integral and related integrals

Theorem 4. If β 1 and α< 2β + 1, then the function ≥ 1 x1−α λx2β−α+1 α 1 α 1 3 λ2x2β G(x)= F 1, + + 1; + + 2, , 2; , λ 1 α − 2β α + 1 2 3 −2β 2β −2β 2β 2 − 4 − − where 2F3 is a hypergeometric function [1] and λ is an arbitrarily constant, is the antiderivative of cos (λxβ) the function g(x)= . Thus, λxα cos (λxβ) 1 x1−α 1 λx2β−α+1 α 1 α 1 3 λ2x2β dx = F 1, + + 1; + + 2, , 2; + C, λxα λ 1 α − 2 2β α + 1 2 3 −2β 2β −2β 2β 2 − 4 Z − − (3.16) and for x 1, | |≫ λx2β−α+1 α 1 α 1 3 λ2x2β F 1, + + 1; + + 2, , 2; 2β α + 1 2 3 −2β 2β −2β 2β 2 − 4 − (3.17) √πλ α 1 2 −α/β+1/β+2 √π 2 cos(λxβ) Γ + + 1 + x−α+1 + . ∼ 2β −2β 2β λ λβ λ2β xβ+α−1

Proof. If β 1 and α< 2β + 1, ≥ ∞ cos (λxβ) 1 (λxβ)2n g(x)dx = dx = ( 1)n dx λxα λxα − (2n)! Z Z Z n=0 ∞ X 1 1 λ2n = dx + ( 1)n x2βn−αdx λxα λ − (2n)! Z Z n=1 ∞ X 1 x1−α 1 λ2n+2 = ( 1)n x2βn+2β−αdx λ 1 α − λ − (2n + 2)! − n=0 Z ∞ X 1 x1−α λ2n x2βn+2β−α+1 = λ ( 1)n + C λ 1 α − − (2n + 2)! 2βn + 2β α + 1 − n=0 − ∞ X 1 x1−α λx2β−α+1 Γ (n α/(2β) + 1/(2β) + 1) = − ( λ2x2β)n + C λ 1 α − 2β Γ(2n + 3)Γ (n α/(2β) + 1/(2β) + 2) − − n=0 − X∞ n 1 x1−α 1 λx2β−α+1 (1) ( α/(2β) + 1/(2β) + 1) λ2x2β/4 = n − n − + C λ 1 α − 2 2β α + 1 − (3/2) (2) ( α/(2β) + 1/(2β) + 2) n! n=0 n n n − − X − 1 x1−α 1 λx2β−α+1 α 1 α 1 3 λ2x2β = F 1, + + 1; + + 2, , 2; + C. λ 1 α − 2 2β α + 1 2 3 −2β 2β −2β 2β 2 − 4 − − 34 Victor Nijimbere

To prove (3.17), we use the asymptotic expression of F (a , a ; b , b , b ; z) for z 1, where 2 3 1 2 1 2 3 − | | ≫ a , a , b , b and b are constants, and π arg z π . It can be obtained using formulas 16.11.1, 1 2 1 2 3 − ≤ ≤ 16.11.2 and 16.11.8 in [9] and is given by F (a , a ; b , b , b ; z)= 2 3 1 2 1 2 3 − R−1 Γ(b )Γ(b )Γ(b ) (a ) Γ(a a n) ( z)−n 1 2 3 z−a1 1 n 1 − 2 − − + O( z −R) Γ(a ) Γ(b a n)Γ(b a n)Γ(b a n) n! | | 2 (n=0 1 1 2 1 3 1 ) X − − − − − − R−1 Γ(b )Γ(b )Γ(b ) (a ) Γ(a a n) ( z)−n + 1 2 3 z−a2 2 n 2 − 1 − − + O( z −R) Γ(a1) ( Γ(b1 a2 n)Γ(b2 a2 n)Γ(b3 a2 n) n! | | ) nX=0 − − − − − − 1/2 −iπ/2 S−1 Γ(b )Γ(b )Γ(b ) e2z e (ze−iπ)(a1+a2−b1−b2−b3+1/2)/2 µ + 1 2 3 n (ze−iπ)−n + O( z −S) Γ(a )Γ(a ) √π 2n+1 | | 1 2 (n=0 ) X 1/2 iπ/2 S−1 Γ(b )Γ(b )Γ(b ) e2z e (zeiπ)(a1+a2−b1−b2−b3+1/2)/2 µ + 1 2 3 n (zeiπ)−n + O( z −S) , Γ(a )Γ(a ) √π 2n+1 | | 1 2 (n=0 ) X (3.18) where the coeﬃcient µn is given by formula 16.11.4 in [9]. λ2x2β α 1 α 1 3 We now set z = , a = 1, a = + + 1, b = + + 2, b = and b =2 in 4 1 2 −2β 2β 1 −2β 2β 2 2 3 (3.18) to obtain α 1 α 1 3 λ2x2β F 1, + + 1; + + 2, , 2; 2 3 −2β 2β −2β 2β 2 − 4 √π α 1 1 √π α 1 2 −α/β+1/β+2 + + 2 + Γ + + 2 (3.19) ∼ λ2 −β β x2β 2 −2β 2β λxβ 2 α 1 cos(λxβ) + + + 2 . λ3 −β β x3β Hence, multiplying (3.19) with λx2β−α+1/(2β α + 1) gives (3.17). − On the other hand, we can show as above that if β 1 and α< 2β + 1, then ≥ cosh (λxβ) 1 x1−α 1 λx2β−α+1 α 1 α 1 3 λ2x2β dx = + F 1, + + 1; + + 2, , 2; +C. λxα λ 1 α 2 2β α + 1 2 3 −2β 2β −2β 2β 2 4 Z − − Theorem 5. Let β 1, then the function ≥ 2 1 λ λ λxβ 5 λ2x2β G(x)= ln x + F 1, 1; 2, , 3; , −2λβx2β − 2 | | 6β 4 2 3 2 − 4 where 2F3 is a hypergeometric function [1] and λ is an arbitrarily constant, is the antiderivative of cos (λxβ) the function g(x)= . Thus, λx2β+1 2 cos (λxβ) 1 λ λ λxβ 5 λ2x2β Ci = dx = ln x + F 1, 1; 2, , 3; + C. β,2β+1 λx2β+1 −2λβx2β − 2 | | 6β 4 2 3 2 − 4 Z (3.20) We also have, cos (λxβ) 1 λx2β 3 λ2x2β Ci = dx = ln x F 1, 1; , 2, 2; + C. (3.21) β,1 λx λ | |− 4β 2 3 2 − 4 Z Some non-elementary integrals of sine, cosine and exponential integrals type 35

Proof.

∞ cos (λxβ) 1 (λxβ)2n Ci = g(x)dx = dx = ( 1)n dx β,2β+1 λx2β+1 λx2β+1 − (2n)! Z Z Z n=0 ∞ X 1 1 λ2n = dx + ( 1)n x2βn−2β−1dx λx2β+1 λ − (2n)! Z Z n=1 ∞ X 1 1 λ2n+2 = ( 1)n x2βn−1dx −2λβx2β − λ − (2n + 2)! n=0 Z X ∞ 1 λ dx λ2n = λ ( 1)n x2βn−1dx −2λβx2β − 2 x − − (2n + 2)! Z n=1 Z ∞X 1 λ λ2n = ln x + λ3 ( 1)n x2βn+2β−1dx −2λβx2β − 2 | | − (2n + 4)! n=0 Z X∞ 1 λ λ2n x2βn+2β = ln x + λ3 ( 1)n + C −2λβx2β − 2 | | − (2n + 4)! 2βn + 2β n=0 X∞ 1 λ λ3x2β (Γ (n + 1))2 ( λ2x2β)n = ln x − + C −2λβx2β − 2 | |− 2β Γ(2n + 5)Γ (n + 2) n! n=0 X2 ∞ n 1 λ λ λxβ (1) (1) λ2x2β/4 = ln x + n n − + C −2λβx2β − 2 | | 6β 4 (2) (5/2) (3) n! n=0 n n n X2 1 λ λ λxβ 5 λ2x2β = ln x + F 1, 1; 2, , 3; + C. −2λβx2β − 2 | | 6β 4 2 3 2 − 4 The proof of (3.21) is similar, we do not show it here.

4. Evaluation of some integrals involving Siα,β and Ciα,β

cosn (λxβ) The integral dx, where n is a positive integer and β 1,α < 2β + 1, can be λxα ≥ written in terms ofZ (3.16) and then evaluated. cos4 (λxβ) Example 2. In this example, the integral dx is evaluated by linearizing cos4 (λxβ). λxα R cos4 (λxβ) 1 cos (4λxβ) 1 cos (2λxβ) 3 dx = dx + dx + dx = λxα 8 λxα 2 λxα 8 Z Z Z Z 1 x1−α 1 λx2β−α+1 α 1 α 1 3 F 1, + + 1; + + 2, , 2; 4λ2x2β 8λ 1 α − 4 2β α + 1 2 3 −2β 2β −2β 2β 2 − − − 1 x1−α 1 λx2β−α+1 α 1 α 1 3 3x + F 1, + + 1; + + 2, , 2; λ2x2β + +C. 2λ 1 α − 2 2β α + 1 2 3 −2β 2β −2β 2β 2 − 8 − −

sinn (λxβ) If β 1 and α < β + 1, the integral dx, where n is a positive integer, can be written ≥ λxα either in terms of (2.7) if n odd, andR then evaluated. 36 Victor Nijimbere

sin3 (λxβ) Example 3. In this example, the integral dx is evaluated by linearizing sin3 (λxβ). λxα R sin3 (λxβ) 1 sin (3λxβ) 3 sin (λxβ) dx = dx + dx λxα −4 λxα 4 λxα Z Z Z 1 xβ−α+1 α 1 1 α 1 3 3 9λ2x2β = F + + ; + + , ; −4 β α + 1 1 2 −2β 2β 2 −2β 2β 2 2 − 4 − 3 xβ−α+1 α 1 1 α 1 3 3 λ2x2β + F + + ; + + , ; + C. 4 β α + 1 1 2 −2β 2β 2 −2β 2β 2 2 − 4 − Example 4. Let us now evaluate the integrals sin (λ/xµ)dx and cos (λ/xµ)dx. 1. The integral sin (λ/xµ)dx is evaluated usingR the substitutionR u = 1/x and Theorem 1 if µ> 1. Then, we have R λ sin (λuµ) λuµ−1 1 1 1 3 3 λ2u2µ sin dx = du = F + ; + , ; xµ − u2 − µ 1 1 2 −2µ 2 −2µ 2 2 − 4 Z Z − λ (1/x)µ−1 1 1 1 3 3 λ2 = F + ; + , ; + C, µ> 1. − µ 1 1 2 −2µ 2 −2µ 2 2 −4x2µ − (4.22)

The integral sin (λ/xµ)dx is evaluated using the substitution u = 1/x and Theorem 3 if µ = 1. Then, we have R λ sin (λu) (λu/2)2 5 λ2u2 sin dx = du = ln u + F 1, 1; 2, 2, ; x − u2 − | | 6 2 3 2 − 4 Z Z (λ/(2x))2 5 λ2 = ln x + F 1, 1; 2, 2, ; + C. | | 6 2 3 2 −4x2 2. Making the substitution u = 1/x and applying Theorem 4 gives

λ cos (λuµ) 1 λu2µ−1 1 1 3 λ2u2µ cos dx = du = + F 1, + 1; + 2, , 2; xµ − u2 u 2µ 1 2 3 −2µ −2µ 2 − 4 Z Z − λ (1/x)2µ−1 1 1 3 λ2 = x + F 1, + 1; + 2, , 2; + C, µ> 1. 2µ 1 2 3 −2µ −2µ 2 −4x2µ − (4.23)

Making the substitution u = 1/x and applying Theorem 5, then for µ = 1, we have

λ cos (λu) 1 λ λ λu 2 5 λ2x2 cos dx = du = + ln u F 1, 1; 2, , 3; x − u2 2λu2 2 | |− 6 4 2 3 2 − 4 Z Z x2 λ λ λ 2 5 λ2 = ln x F 1, 1; 2, , 3; + C. 2λ − 2 | |− 6 4x 2 3 2 −4x2 5. Evaluation of exponential (Ei) and logarithmic (Li) integrals

Theorem 6. If β 1, then for any constant λ, ≥ β eλx λxβ dx = ln x + F (1, 1; 2, 2; λxβ )+ C, x | | β 2 2 Z Some non-elementary integrals of sine, cosine and exponential integrals type 37 and β eλx λxβ F (1, 1; 2, 2; λxβ ) 2+ , x 1. (5.24) 2 2 ∼− λxβ | |≫

Proof. β ∞ ∞ ∞ eλx 1 (λxβ)n dx λnxβn−1 λn dx = dx = + dx = ln x + xβn−1dx x x n! x n! | | n! Z Z n=0 Z Z n=1 n=1 Z X ∞ X ∞ X λn xβn λn+1 xβn+β = ln x + = ln x + | | n! βn | | (n + 1)! βn + β n=1 n=0 X ∞ X λxβ Γ(n + 1) = ln x + (λxβ)n + C | | β Γ(n + 2)Γ(n + 2) n=0 ∞ X λxβ (1) (1) (λxβ)n λxβ = ln x + n n + C = ln x + F (1, 1; 2, 2; λxβ )+ C. | | β (2) (2) n! | | β 2 2 n=0 n n X To derive the asymptotic expression of λxβ F (1, 1; 2, 2; λxβ ), x 1, we use the asymptotic 2 2 | | ≫ expression of the hypergeometric function F (a , a ; b , b ; z) for z 1, where z C, and 2 2 1 2 1 2 | | ≫ ∈ a1, a2, b1 and b2 are constants. It can be obtained using formulas 16.11.1, 16.11.2 and 16.11.7 in [9] and is given by

2F2 (a1, a2; b1, b2; z)= R−1 Γ(b )Γ(b ) (a ) Γ(a a n) (ze±iπ)−n = 1 2 (ze±iπ)−a1 1 n 1 − 2 − + O( z −R) Γ(a ) Γ(b a n)Γ(b a n) n! | | 2 (n=0 1 1 2 1 n ) X − − − − R−1 Γ(b )Γ(b ) (a ) Γ(a a n) (ze±iπ)−n (5.25) + 1 2 (ze±iπ)−a2 2 n 2 − 1 − + O( z −R) Γ(a1) ( Γ(b1 a2 n)Γ(b2 a2 n)n n! | | ) nX=0 − − − − S−1 Γ(b )Γ(b ) µ + 1 2 ezza1+a2−b1−b2 n z−n + O( z −S ) , Γ(a )Γ(a ) 2n | | 1 2 (n=0 ) X where the coeﬃcient µn is given by formula 16.11.4. And the upper or lower signs are chosen according as z lies in the upper (above the real axis) or lower half-plane (below the real axis). β Setting z = λx , a1 = 1, a2 = 1, b1 = 2 and b2 = 2 in (5.25) yields

β 2 eλx F (1, 1; 2, 2; λxβ ) − + , x 1. 2 2 ∼ λxβ λ2x2β | |≫ Hence, β eλx λxβ F (1, 1; 2, 2; λx) 2+ , x 1. 2 2 ∼− λxβ | |≫ This ends the proof. Example 5. The random attenuation capacity of a channel or fading capacity [11] can now be evaluated in terms of the natural logarithm ln and the hypergeometric function 2F2 as 1 1 C = E[log (1 + P H 2)] = e1/P E ( ) E fading 2 | | ln 2 1,1 ∞ − 1,1 P 1 1 1 = e1/P ln P + F 1, 1; 2, 2; ln 2 P 2 2 −P 38 Victor Nijimbere

λeβx x Example 6. One can now evaluate e dx in terms of 2F2 using the substitution u = e , and obtain R λuβ β βx βx e λu λe eλe dx = du = ln u + F (1, 1; 2, 2; λuβ )+ C = x + F (1, 1; 2, 2; λeβx)+ C. u β 2 2 β 2 2 Z Z Theorem 7. The logarithmic integral is given by

x dt ln x Li = = ln + ln x F (1, 1; 2, 2; ln x) ln µ F (1, 1; 2, 2; ln µ), µ> 1. ln t ln µ 2 2 − 2 2 Zµ

And for x µ, ≫ x dt x ln x Li = + ln 2 ln µ F (1, 1; 2, 2; ln µ). (5.26) ln t ∼ ln x ln µ − − 2 2 Zµ

P r o o f. Making the substitution u = ln x and using (4.22) gives

x ln x dx eu = du = [ln u + u F (1, 1; 2, 2; u)]ln x ln x u 2 2 ln µ Zµ lnZµ ln x = ln + ln x F (1, 1; 2, 2; ln x) ln µ F (1, 1; 2, 2; ln µ). ln µ 2 2 − 2 2

Now setting z = ln x, a1 = 1, a2 = 1, b1 = 2 and b2 = 2 in (5.24) or in (4.23) yields 2 x F (1, 1; 2, 2; ln x) − + , x 1. 2 2 ∼ ln x (ln x)2 ≫ This gives x ln x F (1, 1; 2, 2; ln x) 2+ , x 1. 2 2 ∼− ln x ≫ Hence for x µ, ≫ x dt x ln x Li = + ln 2 ln µ F (1, 1; 2, 2; ln µ). ln t ∼ ln x ln µ − − 2 2 Zµ

We importantly note that Theorem 7 adds the term ln (ln x/ln µ) 2 ln µ F (1, 1; 2, 2; ln µ) − − 2 2 to the known asymptotic expression of the logarithmic integral in mathematical literature, Li x/ln x [1, 9]. And this term is negligible if x O(106) or higher. ∼ ∼ We can now slightly improve the prime number Theorem [3] as following,

Corollary 2. Let π(x) denotes the number of primes small than or equal to x and µ> 1. Then for x µ, ≫ x ln x π(x) ln 2 ln µ F (1, 1; 2, 2; ln µ). − ln x ∼ ln µ − − 2 2 Some non-elementary integrals of sine, cosine and exponential integrals type 39

The proof follows directly from equation (5.26) in Theorem 7. Example 7. One can now evaluate ln (ln x)dx using integration by parts. R 1 ln (ln x)dx = x ln (ln x) dx − ln x Z Z = x ln (ln x) ln (ln x) ln x F (1, 1; 2, 2; ln x)+ C. − − 2 2

Theorem 8. For β 1 and α < β + 1, we have ≥ β eλx 1 x1−α xβ−α+1 α 1 α 1 Ei = dx = + F 1, + + 1; 2, + + 2; λxβ + C, β,α λxα λ 1 α β α + 1 2 2 −β β −β β Z − − and for x 1, | |≫ λxβ−α+1 α 1 α 1 F 1, + + 1; 2, + + 2; λx β α + 1 2 2 −β β −β β − β (5.27) λ α 1 1 −α/β+1/β+1 x−α+1 1 eλx Γ + + 1 + . ∼ β −β β −λ − β λβ xβ+α−1 We also have,

β eλx 1 λxβ Ei = dx = + ln( x )+ F 1, 1; 2, 2; λxβ + C. (5.28) β,β+1 λxβ+1 −βxβ | | 2β 2 2 Z

Proof. β ∞ ∞ eλx 1 (λxβ)n 1 dx 1 (λxβ)n Ei = dx = dx = += dx β,α λxα λxα n! λ xα λxα n! Z Z n=0 Z Z n=1 ∞ X ∞ X 1 x1−α 1 λn 1 x1−α λn xβn+β−α+1 = + xβn−αdx = + + C λ 1 α λ n! λ 1 α (n + 1)! βn + β α + 1 − nX=1 Z − nX=0 − α 1 ∞ Γ n + + 1 1 x1−α xβ−α+1 − β β n = + λxβ + C λ 1 α β α 1 − n=0 Γ(n + 2)Γ n + + 2 X − β β α 1 ∞ (1) + + 1 n 1 x1−α xβ−α+1 n −β β xβ = + n + C λ 1 α β α + 1 α 1 n! − − n=0 (2) + + 2 X n −β β n 1 x1−α xβ−α+1 α 1 α 1 = + F 1, + + 1; 2, + + 2; λxβ + C. λ 1 α β α + 1 2 2 −β β −β β − − α 1 α 1 Now setting a = 1, a = + + 1, b = 2, b = + +2 and z = λxβ in (5.25) gives, 1 2 −β β 1 2 −β β α 1 α 1 F 1, + + 1; 2, + + 2; λxβ 2 2 −β β −β β β (5.29) α 1 1 α 1 1 −α/β+1/β+1 eλx + + 1 +Γ + + 2 + . ∼− −β β λxβ −β β λxβ λ2x2β 40 Victor Nijimbere

λxβ−α+1 Hence, multiplying (5.29) with gives (5.27). The proof of (5.28) is similar to that β α + 1 of (3.20). −

Theorem 9. For any constants α, β and λ,

α 1 1 α 1 3 3 λ2x2β F + + ; + + , ; = 1 2 −2β 2β 2 −2β 2β 2 2 − 4 1 α 1 α 1 α 1 α 1 F 1, + + 1; 2, + + 2; iλxβ + F 1, + + 1; 2, + + 2; iλxβ , 2 2 2 −β β −β β 2 2 −β β −β β − h (5.30)i or

α 1 1 α 1 3 3 λ2x2β F + + ; + + , ; = 1 2 −2β 2β 2 −2β 2β 2 2 4 1 α 1 α 1 α 1 α 1 F 1, + + 1; 2, + + 2; λxβ + F 1, + + 1; 2, + + 2; λxβ . 2 2 2 −β β −β β 2 2 −β β −β β − h (5.31)i

P r o o f. Using Theorem 8, we obtain

β β sin (λxβ) 1 eiλx e−iλx dx = − dx xα 2i xα Z Z 1 λxβ−α+1 α 1 α 1 = F 1, + + 1; 2, + + 2; iλxβ (5.32) 2 β α + 1 2 2 −β β −β β − hα 1 α 1 + F 1, + + 1; 2, + + 2; iλxβ + C. 2 2 −β β −β β − i Hence, comparing (2.7) with (5.32) gives (5.30). Or on the other hand,

β β sinh (λxβ) eλx e−λx λxβ−α+1 2 dx = − dx = xα xα β α + 1× Z Z − α 1 α 1 α 1 α 1 F 1, + + 1; 2, + + 2; λxβ + F 1, + + 1; 2, + + 2; λxβ + C. 2 2 −β β −β β 2 2 −β β −β β − h i(5.33)

Hence, comparing (2.9) with (5.33) gives (5.31).

Theorem 10. For any constants α, β and λ,

ix2β−α+1 α 1 α 1 3 λ2x2β xβ−α+1 F 1, + + 1; + + 2, , 2; = 2β α + 1 2 3 −2β 2β −2β 2β 2 − 4 β α + 1× − − α 1 α 1 α 1 α 1 F 1, + 1; 2, + + 2; iλxβ F 1, + + 1; 2, + + 2; iλxβ . 2 2 −β − β −β β − 2 2 −β β −β β − h (5.34)i Some non-elementary integrals of sine, cosine and exponential integrals type 41

Or,

x2β−α+1 α 1 α 1 3 λ2x2β F 1, + + 1; + + 2, , 2; 2β α + 1 2 3 −2β 2β −2β 2β 2 4 − xβ−α+1 α 1 α 1 = F 1, + 1; 2, + + 2; λxβ (5.35) β α + 1 2 2 −β − β −β β − h α 1 α 1 + F 1, + + 1; 2, + + 2; λxβ . 2 2 −β β −β β − i

We prove Theorem 10 as Theorem 9 using Theorems 4 and 8.

6. Conclusion

Si = [sin (λxβ)/(λxα)]dx, β 1, α β + 1, and Ci = [cos (λxβ)/(λxα)]dx, β 1, β,α ≥ ≤ β,α ≥ α 2β + 1, were expressed in terms of the hypergeometric functions F and F respectively, ≤ R R 1 2 2 3 and their asymptotic expressions for x 1 were obtained (see Theorems 1,2, 3, 4 and 5). Once | | ≫ derived, formulas for the hyperbolic sine and hyperbolic cosine integrals were readily deduced from those of the sine and cosine integrals. β On the other hand, the exponential integral Ei = (eλx /xα)dx, β 1, α β + 1, and the β,α ≥ ≤ logarithmic integral dx/ ln x were expressed in terms of the hypergeometric function F , and their R 2 2 asymptotic expressions for x 1 were also obtained (see Theorems 6, 7 and 8). Therefore, their R | |≫ corresponding deﬁnite integrals can now be evaluated using the FTC rather than using numerical integration. Using the Euler and hyperbolic identities Siβ,α and Ciβ,α were expressed in terms of Eiβ,α. And hence, some expressions of the hypergeometric functions 1F2 and 2F3 in terms of 2F2 were derived (see Theorems 9 and 10). The evaluation of the logarithmic integral dx/ ln x in terms of the function 2F2 and its asymptotic expression 2F2 for x 1 allowed us to add the term ln (ln x/ln µ) 2 ln µ 2F2(1, 1; 2, 2; ln µ), | |≫ R − − x µ> 1, to the known asymptotic expression of the logarithmic integral, which is Li = 2 dt/ ln t x ∼ x/ln x [1, 9], so that it is given by Li = dt/ln t x/ln x+ln (ln x/ln µ) 2 ln µ F (1, 1; 2, 2; ln µ) µ ∼ − − 2 2 R in Theorem 7. Beside, this leads to Corollary 2 which is an improvement of the prime number The- R orem [3]. In addition, other non-elementary integrals which can be written in terms of Eiβ,1 and dx/ ln x βx and then evaluated were given as examples. For instance, using substitution, the eλe dx was R written in terms of Ei and therefore evaluated in terms of F , and using integration by parts, β,1 2 2 R the non-elementary integral ln(ln x)dx was written in terms of dx/ ln x and therefore evaluated in terms of F . 2 2 R R REFERENCES

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